I'm not the best functional equation solver, but here goes: (a) Define $f$ as: $f(x)=\begin{cases}0&{x=0,}\\1&{0<x\leq 1.}\end{cases}$ This function works as long as the arguments are valid. (b) $P(x,0) \Rightarrow f(f(x)) = f(x) + f(0)$. Next, notice that $f^{19}(x) = f(x) + 18f(0)$ and $f^{98}(x) = f(x) + 97f(0)$. Thus, (b) holds if $f(0) = 0$. Assume by contradiction that $f(0) \neq 0$. Notice that $$\begin{matrix}P(0,0) \Rightarrow f(f(0)) = 2f(0)\\ P(0,f(0)) \Rightarrow f(2f(0)) = 3f(0)\\ P(0,2f(0)) \Rightarrow f(3f(0)) = 4f(0)\\ \vdots \\ P(0,(n-1)f(0)) \Rightarrow f(nf(0))=(n+1)f(0). \end{matrix}$$However, there will eventually be a choice for $n$ where $nf(0) \leq 1$ and $(n+1)f(0) > 1$, which contradicts the range of $f$. Hence $f(0) = 0$ and therefore $f^{19}(x)=f^{98}(x)$ for some $x$.